f(x) = (x²-1)|x²-3x+2| +cos(|x|)

 

As most of you have pointed out cos(|x|) is a red herring as cos(|x|) = cosx, so the point x=0 can be ignored.

 

So, the critical points become the roots of x²-3x+2, x=1 and x=2

 

g(x) = (x²-1)|x²-3x+2| can be written out as

 

g(x) =   (x²-1)(x²-3x+2)  -<x1

       = -(x²-1)(x²-3x+2)  1x2

      =  (x²-1)(x²-3x+2)  2x<

 

I agree with all of you when you say that it is not differentiable at x=2. But it is definitely differentiable at x=1.

 

Consider P(x) = (x²-1)(x²-3x+2) = (x-1)²(x+1)(x-2)

 

x=1 is a double root of P(x) and hence also a root of P'(x).

 

That means the slope of g(x) is zero for x1^- and  x1⁺.

 

AT x=2, you get a non-zero value for P'(x). So, the slopes on either side of 2 will be equal in magnitude but opposite in sign. So, it is not differentiable at x=2.